JEE Chemistry Equilibrium MCQs Set 31

Answer: (d) \( K_c = \dfrac{1.36}{0.0821\times499} \)

Explanation:
1. \( \Delta n = +1 \), so \( K_p = K_c(RT) \).
2. \( K_c = K_p/(RT) = 1.36/(0.0821\times499) \).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Answer: (b) \( 5.59\times10^{-3} \)

Explanation:
1. The required reaction is the reverse of half the given one.
2. \( K = (1/3.2\times10^4)^{1/2} = 5.59\times10^{-3} \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Answer: (d) low temperature and high pressure

Explanation:
1. The reaction is exothermic: low temperature.
2. 5 moles of gas give 2, so high pressure.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (b) 194

Explanation:
1. Start with 1 mol monomer: monomer = 0.08, dimer = 0.46, total = 0.54 mol.
2. \( p_{monomer} = 0.0296 \) atm and \( p_{dimer} = 0.1704 \) atm.
3. \( K_p = 0.1704/0.0296^2 = 194 \) atm\(^{-1}\).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Answer: (a) \( \dfrac{1}{20P_0^2} \) and \( \dfrac{3}{20P_0^2} \)

Explanation:
1. Let \( p_{N_2H_4} = x \) and \( p_{N_2} = y \). Total: y + x + P0 + 2P0 = 7P0, so y + x = 4P0.
2. The N : H atom ratio stays 18 : 26: 13(2y + P0 + 2x) = 9(4P0 + 3P0 + 4x), giving 26y - 10x = 50P0.
3. Solving: x = 1.5P0, y = 2.5P0. \( K_{p1} = P_0^2/(2.5P_0\times8P_0^3) = 1/(20P_0^2) \) and \( K_{p2} = 1.5P_0/(2.5P_0\times4P_0^2) = 3/(20P_0^2) \).

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (c) less than 15.2 mm Hg

Explanation:
1. At equilibrium \( p_{H_2O} = 0.02 \) atm = 15.2 mm Hg.
2. If the surrounding vapour pressure is lower, the hydrate loses water (efflorescence).

Teacher's Note:
1. In a heterogeneous equilibrium pure solids and liquids do not appear in K, so Kp is a product of gas partial pressures.
2. Use the stoichiometric ratio of the gases formed to find each partial pressure.

Answer: (c) 0.2 atm

Explanation:
1. The \( \ce{CaCO3} \) equilibrium fixes \( p_{CO_2} = 0.04 \) atm.
2. \( p_{CO}^2 = K_p\,p_{CO_2} = 0.04 \), so \( p_{CO} = 0.2 \) atm.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (d) 100 atm

Explanation:
1. The two equilibria share no common gas, so each exerts its own equilibrium pressure independent of the volume (solids in excess).
2. Total = 40 + 60 = 100 atm.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (d) 74.3%

Explanation:
1. \( p_{H_2O} = 2.23\times10^{-2} \) atm = 16.95 torr.
2. Equilibrium relative humidity = 16.95/22.8 = 74.3%.
3. Below this the hydrate is efflorescent.

Teacher's Note:
1. In a heterogeneous equilibrium pure solids and liquids do not appear in K, so Kp is a product of gas partial pressures.
2. Use the stoichiometric ratio of the gases formed to find each partial pressure.

Answer: (d) 4

Explanation:
1. Concentrations: 0.5, 0.25 and 1.0 mol/L.
2. \( K_c = 1.0/(0.5\times0.25^{1/2}) = 1/(0.5\times0.5) = 4 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Answer: (b) 80 atm

Explanation:
1. Let \( p_D = d \). \( p_B = 625/d \), \( p_E = 975/d \), and \( d = p_B + p_E = 1600/d \), so \( d = 40 \) atm.
2. \( p_B + p_E = 40 \).
3. Total = 40 + 40 = 80 atm.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (b) 60 atm

Explanation:
1. Let \( p_C = c = p_B + p_E \), with \( p_B = 300/c \) and \( p_E = 600/c \).
2. \( c^2 = 900 \), so \( c = 30 \) atm.
3. Total = \( 2c = 60 \) atm.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (a) \( K_1\times K_3 = 1 \)

Explanation:
1. (iii) is the reverse of (i), so \( K_1K_3 = 1 \).
2. \( K_2 = \sqrt{K_1} \) and \( K_4 = 1/K_2 = 1/\sqrt{K_1} \); hence \( K_1K_4 = \sqrt{K_1} \neq 1 \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Answer: (c) A, B and D

Explanation:
1. \( K = (p_{H_2O}/p_{H_2})^2 \): at 900 K it is \( (55/45)^2 = 1.5 \) and at 1100 K it is \( (76/24)^2 = 10 \).
2. K rises with temperature, so the reaction is endothermic and reduction is more efficient at higher temperature (and less so at lower).

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.

Answer: (d) B and C

Explanation:
1. R.H. > 1 means the vapour pressure exceeds the saturation value (Q > K), so vapour condenses: leftward.
2. R.H. < 1 means Q < K, so water evaporates: rightward.

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Answer: (c) A and C

Explanation:
1. Increasing a reactant's pressure and raising the temperature (endothermic) both shift the equilibrium forward.
2. Adding solid carbon has no effect, and reducing the volume shifts the equilibrium backward (1 mole to 2 moles of gas).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (c) \( (a+b) > (c+d) \), \( \Delta H < 0 \)

Explanation:
1. Low pressure favours the side with more moles; since it shifts backward, the reactants side has more moles: \( (a+b) > (c+d) \).
2. High temperature shifts it backward, so the forward reaction is exothermic: \( \Delta H < 0 \).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (c) 15/6 mol

Explanation:
1. \( K_p = p_{BrCl}^2/p_{Cl_2} = 1 \) gives \( p_{Cl_2} = 0.01 \) atm.
2. Gas amounts are in the ratio of partial pressures: \( \ce{Cl2} \) = n, BrCl = 10n, \( \ce{Br2} \) vapour = 10n.
3. Cl balance: n + 5n = 1, so n = 1/6.
4. \( \ce{Br2} \) used to make BrCl = 5/6 mol and \( \ce{Br2} \) in vapour = 10/6 mol; minimum added = 15/6 mol.

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (b) D and B

Explanation:
1. \( K_p/K_c = (RT)^{\Delta n} \) with RT = 24.5 L atm/mol.
2. \( \Delta n \) values: A: +1, B: -1, C: +2, D: +3.
3. The ratio is largest for D (\( \Delta n = +3 \)) and smallest for B (\( \Delta n = -1 \)).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Answer: (a) A, B and C

Explanation:
1. Doubling the volume lowers every concentration and shifts the equilibrium backward.
2. Temperature changes K, and adding \( \ce{O2} \) shifts the equilibrium forward.
3. Helium at constant volume changes no partial pressure, so (D) has no effect.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Answer: (d) 2

Explanation:
1. With 1 mol each, 0.5 mol of ester forms, so \( K = 0.5^2/0.5^2 = 1 \).
2. With x mol acid and 1 mol alcohol, ester = x/3: \( K = (x/3)^2/[(2x/3)(1 - x/3)] = 1 \).
3. This gives \( x/9 = (2/3)(1 - x/3) \), so \( x = 2 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Answer: (d) 38

Explanation:
1. \( p_{CO_2} = 2 \) atm, so \( K_p = 4^2\times2 = 32 \) and the total pressure = 6 atm.
2. With solid still in excess, the equilibrium pressure at the same temperature stays 6 atm, so y = 6.
3. \( x + y = 38 \).

Teacher's Note:
1. In a heterogeneous equilibrium pure solids and liquids do not appear in K, so Kp is a product of gas partial pressures.
2. Use the stoichiometric ratio of the gases formed to find each partial pressure.

Answer: (a) 8

Explanation:
1. \( p_B = 1 \), \( p_C = 1.5 \) (ratio 2 : 3), \( p_D = 6 \).
2. \( K_{p_1} = p_B^2p_C^3 = 3.375 \) and \( K_{p_2} = p_D^3 = 216 \).
3. \( 216/(8\times3.375) = 8 \).

Teacher's Note:
1. For simultaneous equilibria sharing a gas, the common species has one partial pressure that satisfies every expression.
2. Write each Kp, express every unknown through the common pressure and solve.

Answer: (c) 9

Explanation:
1. Acid left = 0.1 L \(\times\) 0.75 = 0.075 mol; ester = water = 0.025 mol; alcohol = 0.075 mol.
2. \( K_c = 0.025^2/0.075^2 = 1/9 \).
3. \( 1/K_c = 9 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Answer: (c) 0.0625

Explanation:
1. The gas from the solid is 1.5 - 1 = 0.5 atm, shared equally: 0.25 atm each.
2. \( K_p = 0.25\times0.25 = 0.0625 \) atm\(^2\).

Teacher's Note:
1. In a heterogeneous equilibrium pure solids and liquids do not appear in K, so Kp is a product of gas partial pressures.
2. Use the stoichiometric ratio of the gases formed to find each partial pressure.